An=n+1,Bn=tan(An)*tan(An+1),求Bn的前n项和

来源:学生作业帮助网 编辑:六六作业网 时间:2024/11/18 08:40:01
An=n+1,Bn=tan(An)*tan(An+1),求Bn的前n项和An=n+1,Bn=tan(An)*tan(An+1),求Bn的前n项和An=n+1,Bn=tan(An)*tan(An+1),

An=n+1,Bn=tan(An)*tan(An+1),求Bn的前n项和
An=n+1,Bn=tan(An)*tan(An+1),求Bn的前n项和

An=n+1,Bn=tan(An)*tan(An+1),求Bn的前n项和
tan[n+2-(n+1)]=tan(1)={tan(n+2)-tan(n+1)}/[1+tan(n+2)tan(n+1)],
tan(n+2)tan(n+1)=b(n)={tan(n+2)-tan(n+1)}/tan(1) - 1
b(1)+b(2)+...+b(n)=[1/tan(1)]{tan(3)-tan(2) + tan(4)-tan(3)+...+tan(n+1)-tan(n) + tan(n+2)-tan(n+1)} - n
=[1/tan(1)]{tan(n+2)-tan(2)} - n
=[tan(n+2)-tan(2)]/tan(1) - n

An=n+1,Bn=tan(An)*tan(An+1),求Bn的前n项和 已知bn=tan an*tan an+1,an=n+1,求数列bn前n项的和 等差数列{an},{bn}的前n项和分别为An,Bn,切An/Bn=2n/3n+1,求lim(n→∞)an/bn 计算等差数列{an}{bn}的前n项和分别为An.Bn,且An/Bn=2n/(n+1)求limn→∞(an/bn) 已知an=n/(2^n),bn=ln(1+an)+1/2 an^2,证明,对一切n∈N*,2/(2+an)<an/bn成立 在数列an中,已知a1=2,an+1=2an/an +1,令bn=an(an -1).求证bn的前n项和 数列b=bn+an,an=1/(2^(n-1)),求bn. 已知数列{an}是等差数列,且bn=an+a(n-1),求证bn也是等差数列 a1=1,a2=2,an+2=(an+an-1)/2,n∈N+,(1)令bn=an+1-an,证明bn是等比数列 an是等差数列,an=13+(n-1)2.an=log2bn.证明bn是等比数列 在数1和100之间插入n个实数,使得这n+2个数构成递增的等比数列,将这n+2个数的乘积记为Tn,再令An=lgTn,n≥1(1)求数列{An}的通项公式;(2)设Bn=tan An·tan An+1,求数列Bn的前n项和Sn.【注】tan An+1指tan A(n+1 求证极限:设数列{An},{Bn}均收敛,An=n(Bn-Bn-1),求证limAn = 0.设数列{An},{Bn}均收敛,An=n(Bn-Bn-1),求证limAn = 0. 已知在直角坐标系中,An(an,0),Bn(0,bn)(n∈N*),其中数列{an},{bn}都是递增数列……已知在直角坐标系中,An(an,0),Bn(0,bn)(n∈N*),其中数列{an},{bn}都是递增数列.(1)若an=2n+1,bn=3n+1,判断直线A1B1与A2B2是否 已知:an+sn=n.1、令bn=an-1,求证:{bn}是等比数列.2、求an 设A1=2,An+1=2/An+1,Bn=|An+2/An-1|,n属于正整数,则数列{Bn}的通项公式Bn= 数学已知{an}中,Sn+an=2 1)求an 2)若{bn}中,b1=1,且b(n+1)=bn+an,求bn 已知数列{an},an=2n+1,数列{bn},bn=1/2^n.求数列{an/bn}的前n项和 设{an}是等差数列,an=2n-1,{bn}是等比数列,bn=2^(n-1)求{an/bn}前n项和Sn